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Pickett

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Posts posted by Pickett


  1. Chuck

    For brevity's sake, let's even ignore the original part of the OP that says "of which one is guilty" and first assume that no one killed Ed.

    This means that everyone's first statement ("I did not kill Ed") is true for everyone. But since none of them killed Ed, Alex's second statement would also be true...which means he told 2 truthful statements...therefore, SOMEONE killed Ed.

    If Alex killed Ed:

    • This means that Alex's first statement is FALSE...which means his second statement must be TRUE. However, his second statement is in fact FALSE, because Bob didn't kill Ed and one of them did. Therefore Alex didn't kill Ed.

    If Bob killed Ed:

    • This means that Alex's first statement is TRUE...which means his second statement must be FALSE. However, it is in fact TRUE, because Bob did kill Ed. Therefore Bob did not kill Ed.

    If Duke killed Ed:

    • Bob's first statement is TRUE...which means his second statement must be FALSE. However, it is in fact TRUE because Duke did kill Ed. Therefore Duke did not kill Ed.

    If Chuck killed Ed:

    • Alex's first statement is TRUE, which means his second statement must be FALSE...this is accurate because Bob didn't kill Ed AND one of them did.
    • Bob's first statement is TRUE, which means his second statement must be FALSE...this is accurate because Duke didn't kill Ed.
    • Chuck's first statement is FALSE, which means his second statement must be TRUE...this is accurate because Bob is lying when he said Duke killed Ed (as previously shown above).
    • Duke's first statement is TRUE, which means his second statement must be FALSE...this is accurate because Alex didn't kill Ed, but neither did Bob...

    Therefore Chuck is the killer.


  2. The vodka drinker is in the yellow house no matter what...



    Quick question for clarification: In the clue "One sometimes smells coffee near the green house"...would you take that to mean that the Green house drinks coffee (since you can smell it when you're near it), or would you take that to mean one house away from the Green house, you can smell coffee?

    If the answer to that question is that the Green house drinks coffee, then the Lizard owner drinks coffee (coincidentally in the green house)...and here's the solution (in the order of the houses):
    BEIGE          GREEN     PINK         BLUE     YELLOW
    OJ             Coffee    Milk         Tea      Vodka
    German         Russian   Canadian     Dane     Norwegian
    Dog            Lizard    Goldfish     Horse    Lion
    Heart Attack   Cigar     Non-smoker   Kool     Chesterfield

    Either way, I think this solution satisfies all of the clues above...


  3. My reasoning was correct, but I just made a silly mistake at the end.

    BTW, I was just kidding about getting respect, but in my solution I was still assuming that the puzzle was indeed unsolvable 4 years ago. ;)

    The ages are 18,16 and 10.

    Most combinations of 3 ages will result in a distinct sum and product making the puzzle solvable given the sum and the product of the ages. I found a total of 88 sets of 3 ages (in the range 1-20) that result in non-unique sum and product. These 88 sets of ages make up 44 set pairs with 44 unique sum and product combinations:

    sum, product:    age set 1;    age set 2;
     13,      36:    6,  6,  1;    9,  2,  2;
     14,      40:    8,  5,  1;   10,  2,  2;
     14,      72:    6,  6,  2;    8,  3,  3;
     16,      90:    9,  5,  2;   10,  3,  3;
     17,     144:    8,  6,  3;    9,  4,  4;
     19,     144:    9,  8,  2;   12,  4,  3;
     20,      90:   10,  9,  1;   15,  3,  2;
     21,      96:   12,  8,  1;   16,  3,  2;
     21,     168:   12,  7,  2;   14,  4,  3;
     21,     240:   10,  8,  3;   12,  5,  4;
     22,     360:    9,  8,  5;   10,  6,  6;
     23,     360:   10,  9,  4;   12,  6,  5;
     24,     240:   12, 10,  2;   16,  5,  3;
     25,     360:   12, 10,  3;   15,  6,  4;
     26,     270:   15,  9,  2;   18,  5,  3;
     26,     288:   12, 12,  2;   18,  4,  4;
     27,     480:   15,  8,  4;   16,  6,  5;
     28,     320:   16, 10,  2;   20,  4,  4;
     28,     432:   16,  9,  3;   18,  6,  4;
     28,     450:   15, 10,  3;   18,  5,  5;
     28,     560:   14, 10,  4;   16,  7,  5;
     28,     576:   12, 12,  4;   16,  6,  6;
     28,     630:   14,  9,  5;   15,  7,  6;
     29,     360:   15, 12,  2;   20,  6,  3;
     29,     504:   14, 12,  3;   18,  7,  4;
     29,     720:   12, 12,  5;   15,  8,  6;
     30,     840:   14, 10,  6;   15,  8,  7;
     31,     720:   15, 12,  4;   18,  8,  5;
     31,    1008:   12, 12,  7;   14,  9,  8;
     32,     720:   18, 10,  4;   20,  6,  6;
     32,    1008:   14, 12,  6;   16,  9,  7;
     33,     840:   15, 14,  4;   20,  7,  6;
     34,     900:   15, 15,  4;   20,  9,  5;
     34,    1152:   16, 12,  6;   18,  8,  8;
     35,    1080:   18, 12,  5;   20,  9,  6;
     35,    1120:   16, 14,  5;   20,  8,  7;
     35,    1260:   15, 14,  6;   18, 10,  7;
     35,    1440:   15, 12,  8;   16, 10,  9;
     36,    1200:   16, 15,  5;   20, 10,  6;
     37,    1440:   16, 15,  6;   20,  9,  8;
     38,    1800:   15, 15,  8;   18, 10, 10;
     40,    2160:   16, 15,  9;   18, 12, 10;
     41,    2160:   18, 15,  8;   20, 12,  9;
     44,    2880:   18, 16, 10;   20, 12, 12;
    

    If we subtract 4 from these sets of ages we will get the ages 4 years ago and most of these sets would result in a solvable puzzle 4 years ago. The only 3 sets of ages that would make this puzzle also unsolvable 4 years ago are these:

    sum, product:     ages now;  4 years ago;
     33,     840:   20,  7,  6;    16,  3, 2;
     35,    1440:   16, 10,  9;    12,  6, 5;
     44,    2880:   18, 16, 10;    14, 12, 6;
    

    Two of the 3 sets have 1 year age difference, so if the answer to the last question was "yes" the puzzle would still not have a unique answer. The only unique answer is possible if the answer was "no".

    Hmm...I must have a flaw in my program as this was the approach I was going with, but I had more than the 44 sets that you came up with (which is why I couldn't narrow it down at the end)...nice solve k-man...I'll have to go back and see where my program is wrong.


  4. June is the only month that, besides the initial letter, doesn't contain any tall letters (h, l, t, d, b, etc) or letters that drop below the writing line (g, y, q, etc)

    or

    March...it's the only word that has an english anagram: charm

    EDIT: nevermind on the anagram...May=yam...


    • We have a 1/infinity chance of firing a bullet at 1 m/s.
    • Even if there are other bullets on the line, if none are going 1m/s, we will EVENTUALLY fire enough bullets to collide with all of them, which leaves us with an empty line again
    • That means we have another 1/infinity chance of firing a bullet at 1 m/s...

    This process will repeat an infinite number of times...which means our probably of getting a bullet to NEVER collide is (1/infinity) * infinity...which is 1. However, this is why I said "Almost Surely" above...as I'm certainly not going to wait around for it to be proven. This is just like the Infinite Monkey Theorem...


  5. Since you are firing an infinite number of bullets...the answer is that it is "infinitely improbable"...the ONLY chance you have of a bullet not colliding with another is that it is the FIRST bullet, and that it has a speed of EXACTLY 1 m/s (or, for fun, let's say 0.99999...

    repeating). The chances of this are 1/infinity...which is mathematically zero. However, there is theoretically a CHANCE that it could happen, so you can't say it is "impossible"

    The chances of ANY number in that distribution are 1/infinity (aka, mathematically zero)...however, whenever you pick a number, you have just made it happen...in statistics, it is completely fair to say something has a 0 probability of happening, and yet not be impossible...

    EDIT: Alright, thinking about this more...this gets very interesting...

    Suppose I fire a bullet and it has speed X...then the second bullet has X+Y...they will collide, which means that if the THIRD bullet wouldn't hit either of those no matter how fast it is going...and if it has a speed of 1m/s, it would fly infinitely far.

    We can still make the argument that the only way a bullet would NEVER get hit is if it is going 1 m/s exactly...but we then have to state that it has to be going exactly 1m/s without any other bullets in front of it. Since we're firing forever, I'm going to go ahead and state that since we're firing an infinite number of bullets, were are Almost Surely going to have this event happen at some point...


  6. I'll try to explain this:



    1) Put 4 pegs in the 4 corners of the square
    2) Put a ring on the neck of the sheep and run a rope through it
    3) Put rings on both ends of that rope.
    4) Run a rope from one corner's peg, through the loop on one end of the rope attached to the sheep, and tie it to the peg on the same side of the square
    5) Repeat step 4 for the other side.

    The result would kind of look like an H, where the sheep can roam freely side to side along the middle rope (since it's looped through his neck ring), and up and down (since the middle rope is attached to the two side ropes). This would allow him to eat everywhere inside of the 4 pegs, but not be able to get outside of them at all...


  7. If you don't count words with double letters (like FLOOR), here are a few I came up with:



    4-letters:
    blow
    blot
    belt
    bent

    cent

    crux

    cloy

    dent

    dirt

    deny

    flow
    flux
    fort

    most

    mops

    ...I could go on...

    5-letters:

    forty

    abhor

    ghost

    dirty

    chint

    bijou

    6-letters:

    almost
    biopsy
    abhors
    ghosty

    chintz

    bijoux

    8-letters!
    aegilops

    • Upvote 1

  8. So, isn't the average always going to be 1/NUMBER_OF_GROUPS? In this case, the "average marker group" is always going to 1/3 (aka 23/69).

    As a result, that part of the equation can effectively be ignored...since it's constant (we'll just factor it in at the end to get the final values)

    This leaves us with ONE_GROUP * INVERSE_OF_ANOTHER_GROUP...

    In order to find the maximum, you would always want the INVERSE to the when the group was of length 1...because that means you would always have NUM_MARKERS/1 as one of the multipliers. Which means you just want to choose the largest group as your first one.

    Therefore, to find the MAXIMUM value of n markers, given m groups, it would be

    GROUP 1: (n - m + 1) / n

    GROUP 2: 1 / n

    ...

    GROUP m: 1 / n

    Your maximum value would be (n - m + 1) / m...which in your case (n=23, m=3), would be 7.

    To find the minimum, you would just do the exact opposite (choose the largest group as your reciprocal, and a group of ONE as your base...which is the same grouping as the maximum...which means your minimum value would be 1 / (m * (n-m+1))...which in your case above would simply be 1/63 = 0.015873...

    So basically, you can always use the grouping I have listed above to get both the maximum and minimum values...


  9. This riddle reminds me very much of one that my dad (who just recently passed away) always used to tell me. You can read his telling of it here, on his blog ("I've Got a Blue Spot" was originally going to be the title of a book he was going to write...but turned into the start of a blog instead...the name coming from the riddle):

    http://www.ivegotabluespot.com/archives/18


  10. But how can you be sure that it is exactly half? I mean there will some lost of precision due to estimation.

    I'd say there's as much loss of precision as there is in pouring from one container to the other (there's always left over water in the original container...even if it's just drops)...

    I fully admit that my method has a number of built in assumptions:

    1. My original assumption that the beakers are perfect cylinders/rectangular prisms
    2. The containers hold EXACTLY the number of gallons represented...not a drop more, nor less...(of course, this causes problems in that the second you MOVE a container to pour into another, you would lose some...unless you had an infinitely steady hand...)
    3. You have a perfect eye and can tell the exact second the bottom corner is visible and stop pouring at exactly that time
    4. I'm treating this as a purely mathematical problem...meaning I'm ignoring the surface tension of water, any distortion caused by the beakers in your reading, etc..

    Definitely not a perfect answer, but I'd say you could get "close enough" with it...much closer than just guessing the amounts :thumbsup:

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